need help again! TRIG FNS AND GRAPHS (1 Viewer)

MinnieMinnie

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I know how to do everything but (c)
The answer is supposed to be 0.52


(a) 1.91
(b) 0.95m
(d) 0.5


How would i work out c?
 

CM_Tutor

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Hey where's my logic wrong here. One period of the graph is 720 degrees. So 0.52x2 = 1.04 for b?
This needs to be done in radians. The value of b gives the shift of an x intercept away from the origin.
 

idkkdi

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This needs to be done in radians. The value of b gives the shift of an x intercept away from the origin.
ah ok, that makes sense. But doesn't it look like its shifted by 104 degrees since it's shifted by 2/3 of the first quarter of a period.
 

CM_Tutor

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The zero point is at

which corresponds to your idea, but it doesn't alter the value of itself which was found when and so was not dependent on .
 

idkkdi

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The zero point is at

which corresponds to your idea, but it doesn't alter the value of itself which was found when and so was not dependent on .
y=sin(nt+b)
= sin(n(t+b/n))

At 0.5, b/n = 2b. Inside the brackets is t+2b, therefore the graph of t+2b changes by 2b meaning that if a graph where n =1 is sketched out, it changes by two b. Now it is a matter of applying the multiple n, which basically horizontally shrinks the graph by a factor of 2, thereby making the change only b. However, as it is horizontally shrunk, it appears as if the same proportion of the curve for one period of the curve has been moved.
 

CM_Tutor

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y=sin(nt+b)
= sin(n(t+b/n))

At 0.5, b/n = 2b. Inside the brackets is t+2b, therefore the graph of t+2b changes by 2b meaning that if a graph where n =1 is sketched out, it changes by two b. Now it is a matter of applying the multiple n, which basically horizontally shrinks the graph by a factor of 2, thereby making the change only b. However, as it is horizontally shrunk, it appears as if the same proportion of the curve for one period of the curve has been moved.
If I take the question as having the function: with and given that at we have then:



The period of the curve is and since



Thus, we have

At 3 am, and the tide is at

And, the height was prior to midnight when:



So, the height was at at 1 h 2 min before midnight, or 10:58 pm.

---

Now, I can do the same process with a model and reach the same conclusions, only with a different value of .

---

If I take the question as having the function: with and given that at we have then:



The period of the curve is and since



from which we can now find :



Thus, we have

At 3 am, and the tide is at

And, the height was before midnight when:



So, the height was at at 1 h 2 min before midnight (to the nearest minute), or at 10:58 pm.

----

By either method:
  • the -intercept before occurs at
  • the -intercept occurs at
  • the amplitude is
  • the period is 12.37 h
because the functions and are the same.

Idkkdi, does this answer your question / resolve your confusion?
 
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CM_Tutor

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View attachment 28412
I know how to do everything but (c)
The answer is supposed to be 0.52


(a) 1.91
(b) 0.95m
(d) 0.5


How would i work out c?
MinnieMinnie, may I ask where this came from?

Everyone should note that the way this curve is drawn, with apparent vertical inflexions at each -intercept, is not how a sine function should be drawn.

making the tangents at the intecepts diagonal.

In this specific case:



and so the -intercepts are close to having a gradient of 1... which is not at all how the diagram has been drawn by the source.
 

MinnieMinnie

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MinnieMinnie, may I ask where this came from?

Everyone should note that the way this curve is drawn, with apparent vertical inflexions at each -intercept, is not how a sine function should be drawn.

making the tangents at the intecepts diagonal.

In this specific case:



and so the -intercepts are close to having a gradient of 1... which is not at all how the diagram has been drawn by the source.
This was on our HW sheet!!! :)
 

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